A particle moves in a straight line under a force such that its displacement
s(t), in metres, at time t seconds, is given by s(t) = t3 − 5t2 + 3t +15
(i) Find the expression for the velocity of the particle.
(ii) Find the time at which the particle is at rest.
(iii) Find the acceleration at time t = 5.
(i) v(t) = ds/dt
(ii) set v=0 and solve for t
(iii) a(t) = dv/dt; find a(5)
Post answer
To find the answers to these questions, we need to use some basic principles of calculus. Let's go step by step:
(i) To find the expression for the velocity of the particle, we need to differentiate the displacement function with respect to time. The derivative of a function gives us the rate of change. In this case, the derivative of the displacement function will give us the velocity function.
Take the derivative of s(t) = t^3 - 5t^2 + 3t + 15 with respect to t:
s'(t) = 3t^2 - 10t + 3
So, the expression for the velocity of the particle is v(t) = 3t^2 - 10t + 3.
(ii) To find the time at which the particle is at rest, we need to find the time(s) when the velocity is equal to zero. In other words, we need to solve the equation v(t) = 0.
Set the velocity function equal to zero and solve for t:
3t^2 - 10t + 3 = 0
To solve this quadratic equation, you can use the quadratic formula:
t = (-b ± sqrt(b^2 - 4ac)) / (2a)
In this case, a = 3, b = -10, and c = 3. Plug these values into the quadratic formula and solve for t.
(iii) To find the acceleration at time t = 5, we need to find the derivative of the velocity function with respect to time. The derivative of velocity gives us the acceleration.
Take the derivative of v(t) = 3t^2 - 10t + 3 with respect to t:
a(t) = 6t - 10
Now substitute t = 5 into the acceleration function:
a(5) = 6(5) - 10 = 30 - 10 = 20
So, the acceleration at time t = 5 is 20 m/s^2.